4 Al + 3 O2 -> 2 Al2O3 How much aluminum would be needed to completely react with 45 grams of Oz?

50.6g Al would be needed

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

4 moles of aluminium react with 3 moles of oxygen to produce 2 moles of Al₂O₃

To solve this question we need to find the moles of 45g of O₂. Then using the chemical equation find the moles of Al and its mass:

Moles O₂ -Molar mass: 32g/mol-:

45g O₂ * (1mol / 32g) = 1.41 moles O₂

Moles Al:

1.41 moles O₂ * (4mol Al / 3mol O₂) = 1.875 moles Al

Mass Al -Molar mass: 26.98g/mol-:

1.875 moles Al * (26.98g / mol) =

50.6g Al would be needed

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.